Answer:
The machine decreases in value by 20% each year.
After 1 year the machine is worth only .8 of its original value.
The machine will only decrease by .2 * (.8 A) in the second year not .2 * A where A is the original amount.
A1 = .8 A
A2 = .8 A1 = .64 A value of machine after 2 years
.8 is the value Barry needs when the machine is new
.8^n never goes to zero
After 10 years value = .8^10 A = .11 A of its original value